Proving $||v + u|| = ||v||+||u|| \iff \exists \alpha > 0: v = \alpha u$ on a real vector space

Question

I was trying to prove the following:

Let $V$ be an inner-product space over $\mathbb R$, and let $||v||=\sqrt{\left<v,v \right>}$ be a norm on $V$. Prove that: $||v + u|| = ||v||+||u|| \iff \exists \alpha > 0: v = \alpha u$

I was easily able to show that $\exists \alpha > 0: v = \alpha u \Rightarrow ||v + u|| = ||v||+||u||$, but I'm having some trouble with proving the opposite, how can I do this? Can someone give me a hint?

P.s: I'm aware the there is another post concerning this topic, but that other post assumes that $V$ is a complex vector space. This post is only concerning real vector spaces.

edit:

I did the following: $||u+v||=||u||+||v|| \iff \sqrt{<u,u>+2<u,v>+<v,v>}=||u||+||v|| \iff ||u||^2 + 2<u,v> + ||v||^2 = ||u||^2+||v||^2+2||u||||v|| \iff <u,v>=||u||||v||$ which means $<u,v> \ge 0$. However, I'm not sure what to do now to prove that this implies that $u=\alpha v$, for $\alpha>0$.

edit: I was able to prove it "going back" on the proof for Cauchy Schwarz inequality. Thank you.

Answer 1

The claim is false in, for example, $L^{1}$. That is if u and v are non-negative and disjoint support.